Point D on side AB of triangle ABC is chosen so that AD = AC. It is known that the angle CAB

Point D on side AB of triangle ABC is chosen so that AD = AC. It is known that the angle CAB = 13 degrees and the angle ACB = 143 degrees. Find the DCB corner.

Since, by condition, AD = AC, the ACD triangle is isosceles with the base CD.

Then the angle ADC = ACD.

The sum of the inner angles of the triangle is 180, then the angle ADC = ACD = (180 – CAB) / 2 = (180 – 13) / 2 = 83.5.

Then the angle DСВ = АСВ – АСD = 143 – 83.5 = 59.5.

Answer: The value of the angle DCB is 59.5.